Answer
$$(\sin\theta-\cos\theta)(\csc\theta+\sec\theta)=\tan\theta-\cot\theta$$
Work Step by Step
$$A=(\sin\theta-\cos\theta)(\csc\theta+\sec\theta)$$
- Reciprocal Identities:
$$\sec\theta=\frac{1}{\cos\theta}$$
$$\csc\theta=\frac{1}{\sin\theta}$$
Replace into $A$:
$$A=(\sin\theta-\cos\theta)(\frac{1}{\sin\theta}+\frac{1}{\cos\theta})$$
$$A=(\sin\theta-\cos\theta)\Big(\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}\Big)$$
$$A=\frac{(\sin\theta-\cos\theta)(\sin\theta+\cos\theta)}{\sin\theta\cos\theta}$$
$$A=\frac{\sin^2\theta-\cos^2\theta}{\sin\theta\cos\theta}$$
(for $(A-B)(A+B)=A^2-B^2$)
Now we can separate the numerator to eliminate the denominator:
$$A=\frac{\sin^2\theta}{\sin\theta\cos\theta}-\frac{\cos^2\theta}{\sin\theta\cos\theta}$$
$$A=\frac{\sin\theta}{\cos\theta}-\frac{\cos\theta}{\sin\theta}$$
- Quotient Identities:
$$\cot\theta=\frac{\cos\theta}{\sin\theta}$$
$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$
Therefore, $$A=\tan\theta-\cot\theta$$
